Aspire Faculty ID #11965 · Topic: NIMCET 2025 · Just now
NIMCET 2025

Let the line $\frac{x}{4}+\frac{y}{2}=1$ meets the x-axis and y-axis at A and B, respectively. M is the midpoint of side AB, and M' is the image of the point M across the line x + y = 1. Let the point P lie on the line x + y = 1 such that the $\Delta$ABP is an isosceles triangle with AP = BP. Then the distance between M' and P is

Solution

A = $(4,0)$, B = $(0,2)$ 
Midpoint of AB: $M=\left(\frac{4+0}{2},\frac{0+2}{2}\right)=(2,1)$ 

Image of $M(2,1)$ across line $x+y=1$: 
$M'=(0,-1)$ 

Since $AP=BP$, point $P$ lies on perpendicular bisector of AB. 

Perpendicular bisector of AB passes through $M(2,1)$ and has slope $2$: 
$y-1=2(x-2)$ 
$y=2x-3$ 
Also $P$ lies on: $x+y=1$ 
So, $x+2x-3=1$ $3x=4$ 
$x=\frac{4}{3}$ $y=1-\frac{4}{3}=-\frac{1}{3}$ 
So, $P=\left(\frac{4}{3},-\frac{1}{3}\right)$ 
Distance between $M'(0,-1)$ and $P$: 
$d=\sqrt{\left(\frac{4}{3}-0\right)^2+\left(-\frac{1}{3}+1\right)^2}$ 
$d=\sqrt{\frac{16}{9}+\frac{4}{9}}$ 
$d=\sqrt{\frac{20}{9}}$ 
$d=\frac{2\sqrt5}{3}$ 
Final Answer: $\boxed{\frac{2\sqrt5}{3}}$
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